Topic: WILL THEIR LOTTO SYSTEM WORK IN THE SIMPLEST CASE? Posted: February 15 2008 at 10:32pm

My first introduction to the people that frequent Lotto forums was back in 2003 when I discovered some were actually giving a few numbers for the next draw in a 6/49 Lotto game. ie from some 14 million possibilities they had narrowed it down to a handful. Needless to say I regarded them as fruitcakes but thankfully they are not taken as seriously by the lesser fruitcakes these days. Some sadly live in a time warp where any crap that piles on their dung heap of ignorance is warmly received and enlightened information which shows it up is ignored. Yes, according to them the Sun still revolves around the Earth.

Generally, Forums on the Web are a mixed bunch and if allowed are usually taken over by a bunch of nincompoops which dissuades any possible serious contributor from participating.

I have considered changing the forum format for LottoPoster.com but decided to leave it as it is ie open to contributions and private mesage communication and without breaking longstanding links. Generally, the idiots know they are not welcome to post rubbish at LottoPoster.com but other intelligent comments or articles are welcome. I guess the point is I am not pandering to the ignorant and I am sticking to my life long held principle of being a searcher after the truth.

From 2011 a post like this one can be viewed at ColinFairbrother.com using ASP.net or at LottoPoster.com using classic ASP and a forum format and both using the same database for the html text.

Most of the predictors confine themselves to the Pick 3 game with a thousand possibilities and if not basing it on the history they roam in other areas of numerology and the occult. They are no different from any other con artist come "sooth sayer" telling fortunes etc. If you are not aware both Gail Howard and Ken Silver are out and out numerologists although you won't find any admission of that in their writings.

A simple test which shows up the ignorance and deceit of these con-artists is to consider a perfectly feasible Pick 6, Pool 7 Lotto game with 7 possibilities. According to the occcultist ramblings of Howard and Silver 1 2 3 4 5 6 is less likely to occur than the other 6 possibilities in a random selection. What utter bumkum and nonsense!

The thought occurred to me for those that use history why stop at 1,000 possibilities as in Pick 3?

At first you may think that Pick 3 for the simplest case would have the least possibilities. If for each position we have 2 possibilities then we have 8 three digit possibilities (2x2x2) as follows with a 1 in 8 chance of getting a straight correct. This is equivalent to tossing 3 coins together giving a double for 75% of the throws and a triple otherwise. As there are only two possibilities (Heads or Tails) there are no unmatched or all digits different in a throw.

0 0 0

0 0 1

0 1 0

1 0 0

1 1 1

1 1 0

1 0 1

0 1 1

However, for Pick 6 we can go even further by having a 6/7 game ie 7 Combinations of 6 integers from a pool of 7 integers with odds of 1 in 7.

01 02 03 04 05 06

01 02 03 04 05 07

01 02 03 04 06 07

01 02 03 05 06 07

01 02 04 05 06 07

01 03 04 05 06 07

02 03 04 05 06 07

Here is a sample 42 draws.

Comb6

01 02 03 04 06 07

01 02 04 05 06 07

02 03 04 05 06 07

01 02 03 05 06 07

01 02 04 05 06 07

01 02 03 04 05 06

01 02 04 05 06 07

01 03 04 05 06 07

01 02 03 05 06 07

01 02 03 04 05 06

01 02 03 04 05 07

01 02 04 05 06 07

01 02 03 04 05 06

01 02 04 05 06 07

01 02 03 05 06 07

01 02 03 05 06 07

01 03 04 05 06 07

01 02 03 04 06 07

01 02 03 04 05 06

02 03 04 05 06 07

01 03 04 05 06 07

01 03 04 05 06 07

01 02 03 04 06 07

02 03 04 05 06 07

01 02 03 04 06 07

02 03 04 05 06 07

01 02 03 04 06 07

01 02 03 04 05 06

01 03 04 05 06 07

01 02 03 04 05 07

01 02 04 05 06 07

01 02 03 04 05 07

01 03 04 05 06 07

01 02 03 04 05 07

01 02 03 04 05 06

01 02 03 05 06 07

01 02 04 05 06 07

01 02 04 05 06 07

01 02 03 05 06 07

01 02 04 05 06 07

01 02 03 04 05 07

01 02 03 04 05 06

Occurrence for each integer is as follows -

One

Cnt

01

38

06

37

05

37

04

36

02

36

07

35

03

33

Which 6 integers for the next draw? I can guarantee that you will always have a minimum of 5 right in any line you choose from the 7 possibilities and of course, you will only always have at most 1 wrong. Maybe this is one of those "nice" (to quote The Professor) Covers C(7,6,5,5,1)=1?

Let's keep the house margin at 14.3% meaning 85.7% is returned or 6 to 1 and work with a stake of $49.00.

So, at a dollar a line if you played all 7 it will cost you $7.00 and you get $6.00 back every draw to be down a $1.00. You will last precisely 41 draws with a balance of $6.00..

Play 6 and get it right and you are break even otherwise down $6. On a losing streak you could last as little as 8 draws..

Play 5 and get it right you are $1 in front otherwise down $5. On a losing streak you could last as little as 9 draws..

Play 4 and more than 50% of the time you will win but spending $4.00 to be $2 ahead doesn't pay for the Play 4 in the next draw..

Play 3 and win and you are able to pay for the next Play 3 but how can you get ahead even in a run?.

Play 2 is probably the way I would play as a win would put me in front by $4 allowing a kitty to build up in a run to give something to ease you through a losing streak..

Play 1 line and get it right you have a 600% return to be $5.00 in front. The minimum you will last is 49 draws but that is unlikely as it would mean a losing streak of 49 draws.

Couldn't be simpler. For these payouts and the sample draws given, if you had played 01 02 04 05 06 07 only each draw it would have cost $42.00 to win $54.00 and be ahead $12.00; for 01 02 03 04 05 06 you would have broke even.

So, what is the next draw? All you have to do is eliminate 1 integer from 7! Should be a piece of cake for the filterologists.

However, examining the draws and simplifying an analysis by using one character to represent each possibility ie

a = 1 2 3 4 5 6

b = 1 2 3 4 5 7

c = 1 2 3 4 6 7

d = 1 2 3 5 6 7

e = 1 2 4 5 6 7

f = 1 3 4 5 6 7

g = 2 3 4 5 6 7

Now, if we designate the previous 7 draws by the notation such as e g d e a e we find that for the 35 where we have 7 previous draws they are all different and they are all different from the previous 3 draws upwards . In fact there are 7^7 = 823,543 possibilities! What possible advantage could be gained from this?

Previous 7 Draws

Cnt

a b e a e d

1

a d e e d e

1

a e d d f c

1

a e f d a b

1

a f b e b f

1

a g f f c g

1

b a d e e d

1

b e a e d d

1

b e b f b a

1

b f b a d e

1

c a f b e b

1

c a g f f c

1

c g c a f b

1

c g c g c a

1

d a b e a e

1

d d f c a g

1

d e a e f d

1

d e e d e b

1

d f c a g f

1

e a e d d f

1

e a e f d a

1

e b f b a d

1

e d d f c a

1

e f d a b e

1

e g d e a e

1

f b a d e e

1

f b e b f b

1

f c a g f f

1

f c g c g c

1

f d a b e a

1

f f c g c g

1

g c a f b e

1

g c g c a f

1

g d e a e f

1

g f f c g c

1

What about just the previous two draws? Once again as you see in the table below there is nothing consistent and therefore nothing to be gained.

Block

Previous 2 Draws

Cnt

a

b e

1

a

d e

1

a

e b

1

a

e d

1

a

e e

1

a

g e

1

b

a d

1

b

e a

1

b

e b

2

c

a e

1

c

a g

1

c

e g

1

c

g c

2

d

a b

1

d

d e

1

d

e a

1

d

e b

1

d

e c

1

d

e e

1

e

a e

2

e

b a

1

e

b e

1

e

d d

1

e

d e

1

e

e d

2

e

g d

1

f

b a

1

f

b e

1

f

c a

1

f

c g

1

f

d a

1

f

e c

1

g

c a

1

g

c g

1

g

d e

1

g

e e

1

For those that like to look for repeating patterns I don't see one for the integer that is missing in each draw. If you worked on eliminating the integer 1 more than the previously excluded integer then for 5 of the 40 plays this would work but your return is only $30 for a $40 outlay. You need to get 7 to be ahead.

Do we have any takers for proving their Pick 6 system works in the simplest 6/7 case? I doubt it as most rely on obfuscation. They overlook that any system that relies on other than guesswork can be programmed to determine its efficacy!

Regards

Colin Fairbrother

ps The simplest case is a standard trouble-shooting or fault-finding technique an area I have been involved in most of my working life. You don't look for the most complicated answer - rather, you start with the most simple. For an electrical machine, "Is it getting power?" The solution may well be as simple as plugging in the cord. In the case of Lotto why consider millions of possibilities if a system can't be made to work for as little as 7?

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